問題

解法1

遞迴
時間複雜度Ｏ(n^n)?

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1.size() != s2.size())
            return false;
        if (s1.size() ==1)
            return s1==s2;
        int num[256]={0};
        for (int i=0; i<s1.size(); ++i)
        {
            ++num[s1[i]];
            --num[s2[i]];
        }
        for
 
 (int i=0; i< 256; ++i)
            if (num[i])
                return false;
        for (int i=1; i<s1.size(); ++i)
        {
            if (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)) ||
                isScramble(s1.substr(0, i), s2.substr(s1.size() -i)) && isScramble(s1.substr(i), s2.substr(0
 
, s1.size()-i)))
                return true;
        }
        return false;
    }
};

解法二

dp 時間複雜度(n^4) 空間複雜度(n^3)

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1.size() != s2.size())
            return false;
        int n = s1.size();
        if (n==0)
            return
 
 true;
        if (n==1)
            return s1 == s2;
        bool dp[n+1][n][n];
        memset(dp, 0, sizeof(dp));
        for (int x =0; x< n; ++x)
            for (int y=0; y<n; ++y)
                if (s1[x] == s2[y])
                    dp[1][x][y] = true;
        for (int len = 2; len <=n; ++len)
            for (int x=0; x+len <=n; ++x)
                for (int y =0; y+len<=n; ++y)
                    for (int len1=1; len1<len; ++len1)
                    {
                        int len2 = len -len1;
                        if (dp[len1][x][y] && dp[len2][x+len1][y+len1] || dp[len1][x][y+len2] && dp[len2][x+len1][y])
                            dp[len][x][y] = true;
                    }
        return dp[n][0][0];
    }
};